A compound is a strong competitive inhibitor with an a = 10. The total enzyme concentration...
A compound is a strong competitive inhibitor with an a = 10. The total enzyme concentration in the reaction is 10 nM. What substrate concentration reduces the velocity to 250 nM/min, knowing that this enzyme is characterized by a specificity constant of 2.1 x 105 M-1s-1 and a kcat of 25/min?
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A compound is a strong competitive inhibitor with an a = 10. The
total enzyme concentration...
An enzyme binds to a competitive inhibitor with Kd = 1.2 × 10-6 M at pH 7.0 and 25°C.
An enzyme binds to a competitive inhibitor with Kd = 1.2 × 10-6 M at pH 7.0 and 25°C.(a)At what inhibitor concentration will 75% of the enzyme be bound to the inhibitor if there is no substrate present? (b) This enzyme has a Km of 4.0 × 10-5 M and a Vmax of 50 μM/s. At a substrate concentration of 3.0 × 10-4 M, calculate (i) the velocity of reaction in the presence of the inhibitor at 4.8 x 10-5 M (ii) the degree of...
Values for an Enzyme and a substrate are: Km=4 uM and kcat=20/min. For an experiment where...
Values for an Enzyme and a substrate are: Km=4 uM and kcat=20/min. For an experiment where [S]=6mM, it was found that Vo=480nM/min. What was the enzyme concentration? Give your answer in nM. Using the same kcat and Km as the previous question, if [Et]=0.5uM gives a Vo=5 uM/min, what wat the [S]? Give your answer in uM. reaction is run with the kcat=20/min and the Km=4uM. Use the enzyme concentration from question 1 above. A very strong inhibitor is added creating...
For her undergraduate project, Jessica studied an enzyme that catalyzes the reaction A B. For substrate...
For her undergraduate project, Jessica studied an enzyme that catalyzes the reaction A B. For substrate A, she determined 30 min that Km 3.0 HM and kcat Jessica graduated and her project has been passed on to you. Unfortunately, Jessica was so busy that she sometimes forgot to record all of the details of an assay in her lab notebook. Your mentor suggests that you try to back calculate some of the missing concentration values. Assume that the enzyme follows...
For her undergraduate project, Jessica studied an enzyme that catalyzes the reaction A↽−−⇀B. For substrate A,...
For her undergraduate project, Jessica studied an enzyme that catalyzes the reaction A↽−−⇀B. For substrate A, she determined that ?m=2.5 μM and ?cat=35 min−1. Jessica graduated and her project has been passed on to you. Unfortunately, Jessica was so busy that she sometimes forgot to record all of the details of an assay in her lab notebook. Your mentor suggests that you try to back calculate some of the missing concentration values. Assume that the enzyme follows Michaelis–Menten kinetics. 1)...
Applying the Michaelis-Menten Equation II Another enzyme is found that catalyzes the reaction A <===> B...
Applying the Michaelis-Menten Equation II Another enzyme is found that catalyzes the reaction A <===> B Researchers find that the K for the substrate A is 4 uM, and the kcat is 20 min^-1 (a) In an experiment, [A] = 6 mM, and the initial velocity, Vo was 480 nM min^-1 What was the [Et] used in the experiment? (b) In another experiment, [Et] 0.5uM, and the measured Vo=5uM min^-1 What was the [A] used in the experiment? ( c)...
An enzyme catalyzes the reaction M N. The enzyme is present at a concentration of 3.5...
An enzyme catalyzes the reaction M N. The enzyme is present at a concentration of 3.5 nM, and the Vmax is 2.9 PM s-. The Km for substrate M is 6.5 MM. Calculate kcat kcat = 1 What values of Vmax and Km would be observed in the presence of sufficient amounts of an uncompetitive inhibitor to generate an a' of 1.3? apparent Vmax = UM S-1 apparent Km = UM
Consider the enzyme-catalyzed reaction with Vmax=164 (μmol/L)min−1 and KM=32μmol/L. Part A If the total enzyme concentration...
Consider the enzyme-catalyzed reaction with Vmax=164 (μmol/L)min−1 and KM=32μmol/L. Part A If the total enzyme concentration was 6 nmol/L, how many molecules of substrate can a molecule of enzyme process in each minute? Express your answer to three significant figures. kcat kcat = 2.73×104 min−1 Part B Calculate kcat/KM for the enzyme reaction.
An enzyme catalyzes the reaction M↽−−⇀N . An enzyme catalyzes the reaction M = N. The...
An enzyme catalyzes the reaction M↽−−⇀N .
An enzyme catalyzes the reaction M = N. The enzyme is present at a concentration of 3.5 nM, and the Vmax is 1.9 uM -. The Km for substrate M is 2.9 uM. Calculate kcat kcat = 542.86 What values of Vmax and Km would be observed in the presence of sufficient amounts of an uncompetitive inhibitor to generate an a' of 1.5? apparent Vmax = 0.526 UM s-1 apparent Km = 1.2
CHEM3250 Assignment-Enzyme Inhibition Consider the data below for an enzyme catalyzed reaction. T...
CHEM3250 Assignment-Enzyme Inhibition Consider the data below for an enzyme catalyzed reaction. The rate of the reaction has been determined with and without an inhibitor. A total concentration of enzyme of 20 uM was used in the experiment. SHOW WORK AND UNITS!!! Without Inhibitor With Inhibitor [substrate] (mM)Rate of formation of te of formation of product product (mM/min) mM/min) 6.67 5.25 0.49 7.04 38.91 1.0 2.2 6.9 41.8 44.0 1.5 3.5 1 a) On the same graph, plot the data...
You have an inhibitor for an enzyme that you are studying. The concentration of inhibitor used is...
You have an inhibitor for an enzyme that you are studying. The concentration of inhibitor used is 5.50 µM. The following data was collected for the non-inhibited reaction as well as the reaction that was inhibited. mmol/(mL min) mmol/(mL min) mM Substrate Vo Substrate Vo + Inhibitor 0.200 5.000 3.751 0.400 7.500 4.998 0.800 10.000 5.995 1.000 10.700 6.173 2.000 12.500 6.807 4.000 13.600 7.143 a. Plot this data using Excel or a graphing program. Make sure you give your graph has a...
For her undergraduate project, Jessica studied an enzyme that catalyzes the reaction A B. For substrate A, she determined 30 min that Km 3.0 HM and kcat Jessica graduated and her project has been passed on to you. Unfortunately, Jessica was so busy that she sometimes forgot to record all of the details of an assay in her lab notebook. Your mentor suggests that you try to back calculate some of the missing concentration values. Assume that the enzyme follows...
An enzyme catalyzes the reaction M N. The enzyme is present at a concentration of 3.5 nM, and the Vmax is 2.9 PM s-. The Km for substrate M is 6.5 MM. Calculate kcat kcat = 1 What values of Vmax and Km would be observed in the presence of sufficient amounts of an uncompetitive inhibitor to generate an a' of 1.3? apparent Vmax = UM S-1 apparent Km = UM
An enzyme catalyzes the reaction M↽−−⇀N .
An enzyme catalyzes the reaction M = N. The enzyme is present at a concentration of 3.5 nM, and the Vmax is 1.9 uM -. The Km for substrate M is 2.9 uM. Calculate kcat kcat = 542.86 What values of Vmax and Km would be observed in the presence of sufficient amounts of an uncompetitive inhibitor to generate an a' of 1.5? apparent Vmax = 0.526 UM s-1 apparent Km = 1.2
CHEM3250 Assignment-Enzyme Inhibition Consider the data below for an enzyme catalyzed reaction. The rate of the reaction has been determined with and without an inhibitor. A total concentration of enzyme of 20 uM was used in the experiment. SHOW WORK AND UNITS!!! Without Inhibitor With Inhibitor [substrate] (mM)Rate of formation of te of formation of product product (mM/min) mM/min) 6.67 5.25 0.49 7.04 38.91 1.0 2.2 6.9 41.8 44.0 1.5 3.5 1 a) On the same graph, plot the data...
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